The Stacks project

Proof. Let $S = R[x_1, \ldots , x_ n]/I$ be a presentation. Pick $h_ j \in R[x_1, \ldots , x_ n]$ mapping to $g_ j$ in $S$. Then $R[x_1, \ldots , x_ n, x_{n + 1}]/(I, x_{n + 1}h_ j - 1)$ is a presentation of $S_{g_ j}$. Hence $I_ j = (I, x_{n + 1}h_ j - 1)$ is a Koszul-regular ideal in $R[x_1, \ldots , x_ n, x_{n + 1}]$. Pick a prime $I \subset \mathfrak q \subset R[x_1, \ldots , x_ n]$. Then $h_ j \not\in \mathfrak q$ for some $j$ and $\mathfrak q_ j = (\mathfrak q, x_{n + 1}h_ j - 1)$ is a prime ideal of $V(I_ j)$ lying over $\mathfrak q$. Pick $f_1, \ldots , f_ r \in I$ which map to a basis of $I/I^2 \otimes \kappa (\mathfrak q)$. Then $x_{n + 1}h_ j - 1, f_1, \ldots , f_ r$ is a sequence of elements of $I_ j$ which map to a basis of $I_ j \otimes \kappa (\mathfrak q_ j)$. By Nakayama's lemma there exists an $h \in R[x_1, \ldots , x_ n, x_{n + 1}]$ such that $(I_ j)_ h$ is generated by $x_{n + 1}h_ j - 1, f_1, \ldots , f_ r$. We may also assume that $(I_ j)_ h$ is generated by a Koszul regular sequence of some length $e$. Looking at the dimension of $I_ j \otimes \kappa (\mathfrak q_ j)$ we see that $e = r + 1$. Hence by Lemma 15.30.15 we see that $x_{n + 1}h_ j - 1, f_1, \ldots , f_ r$ is a Koszul-regular sequence generating $(I_ j)_ h$ for some $h \in R[x_1, \ldots , x_ n, x_{n + 1}]$, $h \not\in \mathfrak q_ j$. By Lemma 15.30.14 we see that $I_{h'}$ is generated by a Koszul-regular sequence for some $h' \in R[x_1, \ldots , x_ n]$, $h' \not\in \mathfrak q$ as desired. $\square$